09P Oscillations NJC problem set 2025

1 09 Oscillation Problem Solutions P1 (a) Only for half the cycle (see notes) (b) Only for half the cycle (see notes) (c) No (always in opposite direction 𝑎 = −𝜔2𝑥) P2 D A – Amplitude is half of 70 cm B – Kinetic energy is max at equilibrium which occurs at T/4 C – Restoring force is maximum at amplitude and zero at equilibrium so restoring force decreases from t = 0 to t = T/4 D – Correct. Kinetic energy and speed highest at equilibrium. P3 A 𝐸 = 1 2 𝑚𝜔2𝑥02 Hence 𝐸 ∝ 𝑥02 New energy is ¾ E, hence new 𝑥0 = √3 2 𝑥 So change in amplitude = (1 − √3 2 ) 𝑥 = 0.134𝑥 P4 a) (i) The radian is the unit of measurement for angles and is the ratio of the arc length subtended to the radius of the circle. (a) (ii) For simple harmonic motion, angular frequency is 2𝜋 times the frequency of oscillations. (b) (i) Total energy = maximum gravitational potential energy from equilibrium position = 𝑚𝑔ℎ = 0.120 × 9.81 × 0.04 = 4.7 × 10−3𝐽 (b) (ii) Total energy = 1 2 𝑚𝜔2𝑥02 = 4.7 × 10−3 1 2 𝑚4𝜋2𝑓2𝑥02 = 4.7 × 10−3 𝑓 = √ 2×4.7×10−3 0.120×4×𝜋2×0.082 = 0.56 𝐻𝑧 (𝑥0 = 0.080 𝑚) P5 (i) At t1 and t2, x = 1.3m from mid tide level, ie. equilibrium 1.3 = 3.7 𝑠𝑖𝑛 𝜔𝑡 By sub 𝜔 = 2𝜋/𝑇 𝜔𝑡1 = 0.35901 rad and 𝜔𝑡2= 2.7826 rad t1 = 2606 s and t2 = 20195 s Hence ∆t = t2 – t1= 17589 s = 18 000 s (2 s.f.) OR 290 min (2 s.f.) (ii) Rate of rising tide and rate of falling tide is the gradient of tangent drawn at t1 and t2 respectively = 3.7𝜔 𝑐𝑜𝑠 𝜔𝑡1

2 = (3.7 × 2𝜋 45600) 𝑐𝑜𝑠 [( 2𝜋 45600) (2606)] = 4.8 × 10-4 m s-1 P6 D GPE should increase linearly with upwards displacement. The extension decreases with upwards displacement so, EPE should decrease. It is a curve as EPE = ½ kx2 KE is maximum at equilibrium and zero at amplitude positions. TE is constant. P7 A Total energy must be constant, also check value of KE is max at centre and equal to 1 2 𝑚𝑣2 = 1.8𝐽. EPE has negative value as it is set to be zero at the center. The spring has extension at the center so at the top where the extension, the EPE must be lower and hence a negative value. P8 (a) (i) t3, t7 (ii) t4, t8 (b) (i) 𝑓 = 4200 60 = 70𝐻𝑧 (ii) 𝑎0 = 𝜔2𝑥0 = 4𝜋2𝑓2𝑥0 = 4𝜋2 × 702 × 0.025 = 4800 𝑚𝑠−2 (c) P9 (a) (i) Simple harmonic motion occurs when the acceleration of an object is directly proportional to its displacement and the acceleration is always opposite in direction to the displacement. (ii)1 The graph shows that displacement can be both positive and negative, hence the mass is oscillating. (ii)2 The graph is not a straight line (curved at one end), hence acceleration is not directly proportional to displacement and the oscillations are not simple harmonic. (b) (i)1 At the upper amplitude position (i)2 When the sand first loses contact, the maximum acceleration is equal to the acceleration due to free fall 𝜔2𝑥0 = 𝑔 (2𝜋 × 13)2𝑥0 = 9.81 B1 straight line through the